Question

1 mole of an ideal gas is compressed isothermally and reversibly from initial pressure \(x\) kPa to final pressure \(2x\) kPa at 300 K. Find the work done \((R = 8.314 \text{J K}^{-1}\text{mol}^{-1})\)

• A. \(1432\) J
• B. \(1865\) J
• C. \(1296\) J
• D. \(1729\) J

Hint:

Use w = nRT ln(P2/P1) for isothermal reversible compression with P2/P1 = 2.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
For an isothermal reversible process on an ideal gas, the work depends only on the temperature and the ratio of pressures. Compression means work is done on the gas, so the work is positive in the convention where work done on the system is positive.

Step 2: Key Formula or Approach:
\[ W = nRT\ln\frac{P_2}{P_1} = 2.303\,nRT\log\frac{P_2}{P_1} \]
Here \(n = 1\), \(T = 300\) K, \(P_2/P_1 = 2x/x = 2\).

Step 3: Detailed Explanation:
\[ W = 2.303 \times 1 \times 8.314 \times 300 \times \log 2 \]
\[ \log 2 = 0.3010 \]
\[ 2.303 \times 8.314 \times 300 = 5744.3 \]
\[ W = 5744.3 \times 0.3010 = 1729 \text{ J} \]
The sign is positive because the gas is compressed. The other values come from wrong arithmetic: 1432 J, 1296 J and 1865 J do not match \(nRT\ln 2\).

Final Answer:
The magnitude of work done is about 1729 J, option (D). \[ \boxed{1729 \text{ J (D)}} \]